DRAG DROP
You are configuring a multi-subnet IPv6 network for a regional office.
The corporate network administrator allocates the 2001:0db8:1234:0800: :/54 address space
for your use.
You need to identify network IDs of the first and last subnets that you will be able to create at
the office.
Which network IDs should you identify?
To answer, drag the appropriate network IDs to the correct subnets. Each network ID may
be used once, more than once, or not at all. You may need to drag the split bar between
panes or scroll to view content.
Answer: 
Explanation:
why i can not get it ?
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And the person that will say “oh it’s really easy” should be killed
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Last subnet is missing :: at the end making us confused
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The colon : is one field of an IPv6 address which in hexadecimal is 0000 or 16 bits of binary is 0000000000000000, so 2 colons :: mean 2 hexadecimal fields, e.g. 0000 0000 in hexadecimal or 0000000000000000 0000000000000000 in binary
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A 54-legth prefix for the allocated address lead to use 0800 (00001000 00000000) as the subnetting byte-pair. The last two bits from this 16-bit will be valued to 00,01, 10 and 11 for the subnets, so the last subnet generated will be 00001011 11111111 (0bff).
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2001:0db8:1234:0800::/64
Start Range: 2001:0db8:1234:0800:0:0:0:0
End Range: 2001:0db8:1234:0800:ffff:ffff:ffff:ffff
2001:0db8:1234:0800::/54
Start Range: 2001:0db8:1234:0800:0:0:0:0
End Range: 2001:0db8:1234:0bff:ffff:ffff:ffff:ffff
The question says that the allocated range is 2001:0db8:1234:0800::/54 so the last 10 bits of the 0800 number can also be used. If we use /64 then the network part will always remain constant for the 4 sections, i.e. 2001:0db8:1234:0800 and subnetting will begin only from the 5 section. Seeing the choices available I guess the answer will include the /54 prefix and 0bff/54 can be the last network ID. The question could have been more clear though.
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Yes that is true
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I was not able to make head or tail of how the answer was being arrived at, so I studied IPv6 and its rules regarding this question.
IPv6 address has 8 fields each represented by hexadecimal notation as, e.g. 0db8
The colon : expresses one field of an IPv6 address which in hexadecimal is 0000 so 2 colons :: mean 2 fields, e.g. 0000 0000 in hexadecimal
The question offers IPv6 network addresses of 4 fields plus another 2 fields of :: with subnet masks of either 56 bit or 64 bit
example 2001:0db8:1234:0800: :/54
or
example 2001:0db8:1234:0800: :/64
The 54 bit subnet mask can be eliminated because it would require to borrow 10 bits from example the 0800 field or 0801 field or 08ff field, etc.
So the subnet addresses that can be used for the answer have to be in the /64 bit subnet mask range
Logically the first 64 bit subnet that can be used would be 0800::/64
The last 64 bit subnet that can be used would be 0bff::/64
Hence the answer is:
First subnet: 2001:0db8:1234:0800::/64
Last subnet: 2001:0db8:1234:0bff::/64
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De Answer by aiotestking self is dead WRONG.
The answer MUST include /54 (it is given in the question!)
So skip all /64 they are misleading!
Now IPv6 just like IPv4 has some logic….
IP address 2001:db8:1234:800::/54
type GLOBAL-UNICAST
network 2001:db8:1234:800::
Prefix length 54
network range 2001:0db8:1234:0800:0000:0000:0000:0000-2001:0db8:1234:0bff:ffff:ffff:ffff:ffff
Or for Short: 2001:db8:1234:800:/54-2001:db8:1234:bff:/54
Use: http://www.gestioip.net/cgi-bin/subnet_calculator.cgi
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B-Art is wrong.
the /54 means look at the first 54 bits of the address for the network portion. The 54th bit is the second bit in the second hex value of the 4th group ie the 8 in 0800
(2001:0db8:1234:0800::).
^
if we expand the 8 to its binary representation it looks like
1000 == 8 in binary
^^
So the subnet can use these last 2 bits of the 8 + the last two hex values of
0800.
^^
If the answer uses /54 that means it ignore the bits after the 54th bit!
let’s break down the answer
2001:db8:1234:0800:/54
2001:db8:1234:0bff:/54
^^
We know that the last two hex values are ignored so they are equivalent to
2001:db8:1234:0800:/54
2001:db8:1234:0b00:/54
^
lets look at the b and 8. In binary they are
1011 = b
1000 = 8
^^
And we know that the last two bits are ignored! so that makes the value equivalent to
2001:db8:1234:0800:/54
2001:db8:1234:0800:/54
The network ID has to be /64 so that its not ignoring the last 10 bits π
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I was asked by a friend to help with this so here is my two cents π
From wikipedia on IPv6 addresses: Unicast and anycast addresses are typically composed of two logical parts: a 64-bit network prefix used for routing, and a 64-bit interface identifier used to identify a host’s network interface.
So any network in this case will have /64 mask.
That means that the first 64 bits designate the networks. Since we have 54 bits that can not be changed (our upstream provider has given us this range) to create our networks we can only use last 64-54=10 bits of the first 64 bits of the IPv6 addresses. These 10 bits fall into the fourth 16 bit segment of the given address, in this case that is 0800 or written in binary:
HEX(0800) = BIN(0000100000000000)
Of these, only 10 last bits can be changed to 1’s so we get:
BIN(0000101111111111) = HEX(0bff)
Now, when we get all of this back to the whole address, we get:
First network: 2001:0db8:1234:0800::/64
Last network: 2001:0db8:1234:0bff::/64
Total number of networks: 2^10=1024 networks.
So the Adam was correct π And yes, there is one “:” missing in the last network answer.
Hope this helped, as I like this site very much and has helped me pass many exams.
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Thanks BudΕΎi π
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NO the answer is still wrong, the /54 determines the prefix and what can be changed just like an IPv4 class B address. /64 is 1 ip address range which then includes the hexadecimal numbers of the /54.
example
IPv4 Class B 172.168.0.0 /16 or 172.168.0.0 sm 255.255.0.0 (11111111.11111111.00000000.00000000)
ok here’s the math
2 to the power of bits borrowed equals networks [2^0=1]
2 to the power of bits remaining off equals valid addresses [2^16=65536]
Valid addresses minus 2 equals usable hosts [65536-2=65534]
IPv6 is given 48 bits which comes from top level ISPs
2001:0db8:1234: = 48 bits so we have 16 bits to play with for subnets
Organizations can chose any range for the prefix id, this is similar to the class B I referenced.
/49 we borrow 1 and have 2^17 available subnets
/50 we borrow 2 and have 2^16 available subnets on so….
/54 we borrow 6 and have 2^12 available subnets but since we borrowed 6 those are preassigned numbers and will not change, only the subnet interface ID changes 0800 0000 – 0bff ffff and since those numbers are predefined we need to show the prefix id at the end of our IP address to stipulate what is viable to change. So all answers must have /54 since that is what you are given.
2nd the :/54 is not missing an extra colon ::= multiple groups of zeros and since our Subnet addresses can be 0bff ffff no group of zero there hence just a single colon signifying the follow numbers would just be host
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I read all of the other explanations and didn’t understand. Your explanation is so simple and easy to understand.
Thank you!
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Very good budzi
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Good job Budzi, the answer is correct and thanks for the simple logic used in your post.
Every one is scared about ipv6 but if you take a closer look it will be easier from an admin perspective fewer things to configure π
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Invalid IPv6 address: 2001:0db8:1234:0bff:/xx.
Valid IPv6 address: 2001:0db8:1234:0bff::/xx.
https://www.ipv6.com/general/ipv6-addressing/
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Thank you Budzi you made it simple
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